# Find the characteristic polynomial of an arbitrary-sized matrix

• Nov 5th 2009, 06:45 PM
Last_Singularity
Find the characteristic polynomial of an arbitrary-sized matrix
Question: Let $A \in M_{n \times n}(R)$ be the matrix defined by $A_{ij} = 1$ for all $i,j$. Find the characteristic polynomial of $A$.

My attempt: Call $A_k$ to be the $k \times k$ matrix of all $1$'s. Then its characteristic polynomial $f(t)_k = (-1)^k [t^k - k t^{k-1}]$. Use proof by induction.

Base case: $A_1 = [1]$. Then its characteristic polynomial $f(t)_1 = 1-t = -t-(-1) = (-1)[t^1 - 1 t^0]$. This checks.

Then for induction step, assume that it holds for $k$ and attempt to prove it to be true for $k+1$. My problem with this was that as I expanded out the determinant along the top row of $A_{k+1} - t I_{k+1}$, things got very messy and I couldn't see a pattern.

Could you give me a hand, please? Thanks!
• Nov 6th 2009, 12:48 AM
Opalg
One way of doing this is to think in terms of eigenvalues and eigenvectors. It's easy to see that $v = (1,1,...,1)^{\textsc t}$ is an eigenvector of A with eigenvalue n. Also, if w is any vector orthogonal to v then $Aw=0$. So 0 is an eigenvalue whose eigenspace has dimension n–1. Therefore the characteristic polynomial of A must be a scalar multiple of $(t-1)t^{n-1}$. But the coefficient of $t^n$ has to be $(-1)^n$. That gives $f_n(t) = (-1)^n(t^n-nt^{n-1})$, as you wanted.